A loop is formed by two parallel conductors connected by a solenoid with inductance L=2H and a conducting rod of mass m=80g, which can freely (without fricition) slide over the conductors. The conductors are located in a horizontal plane in a uniform vertical magnetic field with induction B=4T.
The distance between the conductors is equal to l=4cm. At the moment t=0 the rod is imparted an initial velocity `v_(0)=2 cm//s` directed to the right. Find the amplitude (in cm) for the oscillations of conducting rod, if the electric resistance of the loop is negligible.

Options

A. 0.25 cm
B. 5 cm
C. 0.5 cm
D. 2.5 cm

Answer

B

Explanation

According to Len's law, magnetic force, will oppose the motion of conducting rod.

here, `Bvl=L(di)/(dt)rArrB(vdt)l=L(di)rArrBl`
`(dx)=L(di)rArr x=(Li)/(Bl)`
Force `=ilB=(B^(2)l^(2))/(L)x=momega^(2)xrArromega=(Bl)/sqrt(mL)`,
So amplitude `=(V_(0))/(omega)=(v_(0))/(Bl)sqrt(mL)=5cm`